Question

A 25.0g sample of an unknown metal was heated to 100.0 degrees C and placed into a beaker containing 90.0g of water at 25.32 degrees C. The temperature of the water rose to a final value of 27.18 degrees C. Neglecting heat loss to the room and the heat capacity of the beaker itself, what is the specific heat of the metal? Using the given specific heat values below (in cal/g degrees C), identify the unknown metal.

Barium 0.068

Chromium 0.111

Copper 0.092

Gold 0.031

Tin 0.051

Rubidium 0.086

Titanium 0.142

Answer 1

m(water) = 90.0 g

T(water) = 25.32 oC

C(water) = 4.184 J/goC

m(metall) = 25.0 g

T(metall) = 100.0 oC

C(metall) = to be calculated

We will be using heat conservation equation

use:

heat lost by metall = heat gained by water

m(metall)*C(metall)*(T(metall)-T) = m(water)*C(water)*(T-T(water))

25.0*C(metall)*(100.0-27.18) = 90.0*4.184*(27.18-25.32)

1820.5*C(metall) = 700.4016

C(metall)= 0.385 J/goC

This is specific heat capacity of gold